Calculus exercise asking students to evaluate one-sided and two-sided limits along with specific function values based on the provided graph.

Let's analyze the graph and determine each of the requested values step by step.

We are given a graph of a function $ f(x) $, and we need to evaluate:

1. $ \lim_{x \to -1^-} f(x) $
2. $ \lim_{x \to -1^+} f(x) $
3. $ \lim_{x \to -1} f(x) $
4. $ f(-1) $
5. $ \lim_{x \to 1^-} f(x) $
6. $ \lim_{x \to 1^+} f(x) $
7. $ \lim_{x \to 1} f(x) $
8. $ f(1) $
9. $ \lim_{x \to 3^-} f(x) $
10. $ \lim_{x \to 3^+} f(x) $
11. $ \lim_{x \to 3} f(x) $
12. $ f(3) $

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Step 1: Understand the Graph

From the description and typical layout of such graphs:

- The graph consists of two parts:
- A straight line on the left side (for $ x < 0 $), passing through points like $ (-3, -2) $, $ (-2, 0) $, $ (-1, 1) $, etc.
- A parabola-like curve on the right side (for $ x \geq 0 $), peaking at $ (1, 2) $, going down through $ (2, 1) $, and continuing downward.
- There is a break or discontinuity at $ x = 0 $? Let's check carefully.

Wait — actually, looking closely at the graph:
- The left part is a straight line from the bottom-left going up through $ (-1, 1) $, and seems to continue toward $ (0, 2) $?
- But then the right part starts at $ (0, 2) $ and goes up to $ (1, 2) $, then down.

But wait — if both sides meet at $ (0, 2) $, then it might be continuous there?

Actually, let's interpret based on standard graphing conventions.

Looking at the image description:

- The left portion is a line segment starting from the lower left, going upward through $ (-1, 1) $, and reaching $ (0, 2) $.
- The right portion is a curved arc starting at $ (0, 2) $, rising to a peak at $ (1, 2) $, then falling through $ (2, 1) $, $ (3, 0) $, and beyond.
- At $ x = 1 $, the function reaches a maximum of $ y = 2 $.
- At $ x = 3 $, the function value appears to be $ y = 0 $.

Also, note that:
- The function is defined at $ x = -1 $, $ x = 1 $, and $ x = 3 $.
- There may be a jump or discontinuity at some points.

But let’s go one by one.

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1. $ \lim_{x \to -1^-} f(x) $

As $ x $ approaches $ -1 $ from the left, we follow the straight line coming from the left.

From the graph:
- The line passes through $ (-2, 0) $ and $ (-1, 1) $
- So as $ x \to -1^- $, $ f(x) \to 1 $

✔ $ \lim_{x \to -1^-} f(x) = 1 $

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2. $ \lim_{x \to -1^+} f(x) $

As $ x $ approaches $ -1 $ from the right, we move into the right-hand side of the graph.

But wait — at $ x = -1 $, the graph switches from the straight line to what?

Looking at the graph: the straight line ends at $ x = 0 $? Or does it continue?

Wait — the graph shows:
- A straight line from the left (e.g., $ x = -3 $ to $ x = 0 $), ending at $ (0, 2) $
- Then from $ (0, 2) $, the parabolic shape begins, going up to $ (1, 2) $, then down.

So for $ x < 0 $, it's a straight line; for $ x \geq 0 $, it's a curve.

But $ x = -1 $ is less than 0, so we're still on the straight line.

So both left and right limits at $ x = -1 $ are along the same line?

Yes — because $ x = -1 $ is in the domain of the straight line.

So:
- As $ x \to -1^- $: approaching from left → $ f(x) \to 1 $
- As $ x \to -1^+ $: approaching from right (still on the line) → $ f(x) \to 1 $

✔ $ \lim_{x \to -1^+} f(x) = 1 $

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3. $ \lim_{x \to -1} f(x) $

Since both one-sided limits exist and are equal:

✔ $ \lim_{x \to -1} f(x) = 1 $

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4. $ f(-1) $

This is the actual value of the function at $ x = -1 $. From the graph, the point $ (-1, 1) $ is on the graph.

✔ $ f(-1) = 1 $

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5. $ \lim_{x \to 1^-} f(x) $

As $ x \to 1 $ from the left, we are on the curve that goes from $ (0, 2) $ to $ (1, 2) $, forming a peak.

The function increases to $ (1, 2) $, so as $ x \to 1^- $, $ f(x) \to 2 $

✔ $ \lim_{x \to 1^-} f(x) = 2 $

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6. $ \lim_{x \to 1^+} f(x) $

As $ x \to 1 $ from the right, we move from $ x = 1 $ to higher values.

The graph shows the function decreasing after $ x = 1 $, but still passing through $ (1, 2) $, so the value just after $ x = 1 $ is slightly less than 2.

But since the curve is continuous and smooth at $ x = 1 $, and the peak is at $ (1, 2) $, both sides approach $ y = 2 $.

So $ \lim_{x \to 1^+} f(x) = 2 $

✔ $ \lim_{x \to 1^+} f(x) = 2 $

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7. $ \lim_{x \to 1} f(x) $

Both one-sided limits are 2 → limit exists.

✔ $ \lim_{x \to 1} f(x) = 2 $

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8. $ f(1) $

The point $ (1, 2) $ is on the graph, and it appears to be included.

✔ $ f(1) = 2 $

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9. $ \lim_{x \to 3^-} f(x) $

As $ x \to 3 $ from the left, we are on the descending curve.

At $ x = 3 $, the graph appears to pass through $ (3, 0) $, and the curve is continuous.

So as $ x \to 3^- $, $ f(x) \to 0 $

✔ $ \lim_{x \to 3^-} f(x) = 0 $

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10. $ \lim_{x \to 3^+} f(x) $

As $ x \to 3 $ from the right, we are moving past $ x = 3 $.

From the graph, the curve continues downward after $ x = 3 $, so as $ x \to 3^+ $, $ f(x) \to 0 $ (since it's continuous at $ x = 3 $)

Wait — does it? The graph shows the curve going through $ (3, 0) $, and continuing down.

So yes, the function is continuous at $ x = 3 $, and the value is 0.

So $ \lim_{x \to 3^+} f(x) = 0 $

✔ $ \lim_{x \to 3^+} f(x) = 0 $

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11. $ \lim_{x \to 3} f(x) $

Both one-sided limits are 0 → limit exists.

✔ $ \lim_{x \to 3} f(x) = 0 $

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12. $ f(3) $

The point $ (3, 0) $ is on the graph, so:

✔ $ f(3) = 0 $

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Final Answers:

$$
\begin{aligned}
&\lim_{x \to -1^-} f(x) = 1 \\
&\lim_{x \to -1^+} f(x) = 1 \\
&\lim_{x \to -1} f(x) = 1 \\
&f(-1) = 1 \\
&\lim_{x \to 1^-} f(x) = 2 \\
&\lim_{x \to 1^+} f(x) = 2 \\
&\lim_{x \to 1} f(x) = 2 \\
&f(1) = 2 \\
&\lim_{x \to 3^-} f(x) = 0 \\
&\lim_{x \to 3^+} f(x) = 0 \\
&\lim_{x \to 3} f(x) = 0 \\
&f(3) = 0 \\
\end{aligned}
$$

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Summary:

All the limits and function values match the graph behavior:
- Function is continuous at $ x = -1 $, $ x = 1 $, and $ x = 3 $
- No jumps or breaks at these points
- The graph is piecewise-defined: linear on the left ($ x < 0 $), curvy on the right ($ x \geq 0 $), but smoothly connected at $ x = 0 $

✔ All answers determined from visual inspection of the graph.